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Math Help - Find the Galois group?

  1. #1
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    Find the Galois group?

    I have the polynomail f(x)=x^5 +2x+2 over rationals. Let M be teh splitting field of this over rationals.

    Then I want to show that the galois group is S_5.

    My argument is as follows:

    f(x) is irreducible by Eisenstein with p=2. and the galois group is isomorphic to a subgroup of S_5, let us call it H.

    we have that the order of H is divisible by 5, then by using Cauchy's theorem we have that there is an element of order 5 in H, so it must be a 5-cycle. So H contains a five cycle.

    Now I can show that f has only one real root?

    Then what will be the argument be? I want to conclude something like that H contains a transposition, so H must be S_5.

    Any help, any hints? Thanks.
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  2. #2
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    Quote Originally Posted by ZetaX View Post
    I have the polynomail f(x)=x^5 +2x+2 over rationals. Let M be teh splitting field of this over rationals.

    Then I want to show that the galois group is S_5.

    My argument is as follows:

    f(x) is irreducible by Eisenstein with p=2. and the galois group is isomorphic to a subgroup of S_5, let us call it H.

    we have that the order of H is divisible by 5, then by using Cauchy's theorem we have that there is an element of order 5 in H, so it must be a 5-cycle. So H contains a five cycle.

    Now I can show that f has only one real root?

    Then what will be the argument be? I want to conclude something like that H contains a transposition, so H must be S_5.

    Any help, any hints? Thanks.
    Modulo 29 this polynomial factors into irreducibles as (x^2+4x+7)(x^3+25x^2+9x+21).
    By Dedekind's theorem its Galois group has a permuation that is a product of a 2-cycle and a 3-cycle.
    The only transitive subgroup of S_5 with this property is S_5 itself.
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  3. #3
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    Thanks for your help, but what abiut the way I started with, doesnot it work in this case?

    How can i show that the only transitive subgroup of S_5 contaning a 3 and 2 cycle is S_5?
    Last edited by ZetaX; March 22nd 2009 at 11:27 PM.
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